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(B) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $m = \pi r^2 \rho$ is the mass per unit le

Vedclass · Accepted Answer

$1: 3$

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(B) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $m = \pi r^2 \rho$ is the mass per unit length.Thus,$n = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.The frequency of the $p$-th overtone is $(p+1)n$.For string $A$,the first overtone is $2n_A = \frac{2}{2l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.For string $B$,the second overtone is $3n_B = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.Given $2n_A = 3n_B$ and $r_A = 2r_B$:$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B}$$\frac{l_A}{l_B} = \frac{2 r_B}{3 r_A} = \frac{2 r_B}{3(2 r_B)} = \frac{2}{6} = \frac{1}{3}$.Therefore,the ratio of the lengths $l_A : l_B = 1 : 3$.

Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$,the ratio of the lengths of the strings is :

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